*ptr1=*ptr2=string;
for(i=0;i<strlen(string);i++)
if(*str==' ')
ptr2++;
else
{
*ptr1=*ptr2;
ptr1++;
ptr2++;
}
hey guys if anything wrong in this code pls let me know............
On Tue, Oct 11, 2011 at 11:04 AM, DIPANKAR DUTTA <[email protected]
> wrote:
> Solution 1:
> need two scan :
>
> start from left and replace one by one by non-space character ( thus
> have minimal replacemnt)
>
> count =0;
> for (int i=0;i<len(str);i++)
> {
> if(str[i] !=NONSPCECHAR)
> { str[count++]=str[i]
> }
>
> }
>
> ps: any way it needs Left shift...and all solution must need left
> shift if you try to compact it in lower index area.
>
>
>
> On 10/11/11, abhishek sharma <[email protected]> wrote:
> > Can in place compaction be done without left shifts?
> >
> >
> >
> > --
> > Nice Day
> >
> > Abhishek Sharma
> > Bachelor of Technology
> > IIT Kanpur (2009)
> >
> > --
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> >
> >
>
>
> --
> Thanks and Regards,
> ------------------------------
> **DIPANKAR DUTTA
> Software Development Engineer
> Xen Server - OpenStack Development Team (DataCenter and Cloud)
>
> Citrix R&D India Pvt Ltd
> 69/3, Millers Road, Bangalore – 560052
> Phone: +91 8147830733
> Office: Extn: 16429
> Email: [email protected]
>
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Thanks & Regards
Prasad Y.Jondhale
M.Tech(software engg.),
Delhi College Of Engineering,
Main Bawana road,Delhi-110042
Ph-09540208001
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